Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{y - 7}{y^2 - 2y - 35} \div \dfrac{y + 7}{y + 5} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{y - 7}{y^2 - 2y - 35} \times \dfrac{y + 5}{y + 7} $ First factor the quadratic. $t = \dfrac{y - 7}{(y + 5)(y - 7)} \times \dfrac{y + 5}{y + 7} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (y - 7) \times (y + 5) } { (y + 5)(y - 7) \times (y + 7) } $ $t = \dfrac{ (y - 7)(y + 5)}{ (y + 5)(y - 7)(y + 7)} $ Notice that $(y - 7)$ and $(y + 5)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ (y - 7)\cancel{(y + 5)}}{ \cancel{(y + 5)}(y - 7)(y + 7)} $ We are dividing by $y + 5$ , so $y + 5 \neq 0$ Therefore, $y \neq -5$ $t = \dfrac{ \cancel{(y - 7)}\cancel{(y + 5)}}{ \cancel{(y + 5)}\cancel{(y - 7)}(y + 7)} $ We are dividing by $y - 7$ , so $y - 7 \neq 0$ Therefore, $y \neq 7$ $t = \dfrac{1}{y + 7} ; \space y \neq -5 ; \space y \neq 7 $